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Here's what comes to mind

In case of independent random variables [itex]E(XY) = E(X)E(Y)[/itex]. By first expanding the expression inside expected value:

[tex]

\begin{align*}

&[(W_u-W_v)^2 + 2(W_u-W_v)W_v + W_v^2]W_v^2 \\

=&(W_u-W_v)^2W_v^2 + 2(W_uW_v -W_v^2)W_v^2 + W_v^4 \\

=&(W_u-W_v)^2W_v^2 + 2W_uW_v^3 - W_v^4

\end{align*}

[/tex]

To get the mgf of Brownian, one computes

[tex]

M_W(x) = \mathbb E(e^{xW_s}) = \mbox{exp}\left ( \frac{1}{2}x^2s\right )

[/tex]

Fourth derivative (w.r.t ##x##) at [itex]x=0[/itex] gives us the fourth moment, which is (if I calculated correctly) [itex]\mathbb E(W_v^4) = 3v^2[/itex].

Since [itex]W_u-W_v, W_v[/itex] are independent, squaring them preserves independence, so

[tex]

\mathbb E(W_u-W_v)^2W_v^2 = \mathbb E(W_u-W_v)^2\mathbb E(W_v^2) = (u-v)v

[/tex]

Not sure what is happening with the middle part, though. Does it vanish?

In case of independent random variables [itex]E(XY) = E(X)E(Y)[/itex]. By first expanding the expression inside expected value:

[tex]

\begin{align*}

&[(W_u-W_v)^2 + 2(W_u-W_v)W_v + W_v^2]W_v^2 \\

=&(W_u-W_v)^2W_v^2 + 2(W_uW_v -W_v^2)W_v^2 + W_v^4 \\

=&(W_u-W_v)^2W_v^2 + 2W_uW_v^3 - W_v^4

\end{align*}

[/tex]

To get the mgf of Brownian, one computes

[tex]

M_W(x) = \mathbb E(e^{xW_s}) = \mbox{exp}\left ( \frac{1}{2}x^2s\right )

[/tex]

Fourth derivative (w.r.t ##x##) at [itex]x=0[/itex] gives us the fourth moment, which is (if I calculated correctly) [itex]\mathbb E(W_v^4) = 3v^2[/itex].

Since [itex]W_u-W_v, W_v[/itex] are independent, squaring them preserves independence, so

[tex]

\mathbb E(W_u-W_v)^2W_v^2 = \mathbb E(W_u-W_v)^2\mathbb E(W_v^2) = (u-v)v

[/tex]

Not sure what is happening with the middle part, though. Does it vanish?

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